Hard GCSE Gradients and areas under curves Questions

Challenging, exam-style GCSE Gradients and areas under curves questions with worked solutions. Stretch yourself on the hardest trapezium rule, 4 strips, convex curve, concave curve problems.

trapezium rule4 stripsconvex curveconcave curveunder-estimateexponential curve
GCSE Higher34 questionsStep-by-step solutions
Question 1
6 markschallenging
A lorry's speed, vv m/s, after tt seconds is modelled by v=40−t2v = 40 - t^2 for 0≤t≤50 \le t \le 5. Use the trapezium rule with 55 strips of width 11 to estimate the distance travelled in the first 55 seconds.
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Worked solution

  1. Write down the trapezium rule

    A≈h2[y0+y5+2(y1+⋯+y4)]A \approx \frac{h}{2}\Big[y_0 + y_{5} + 2\big(y_1 + \dots + y_{4}\big)\Big]

    Add the first and last ordinates, add twice every ordinate in between, then multiply by half the strip width.

  2. Find the strip width

    h=5−05=1h = \frac{5 - 0}{5} = 1

    The interval from t=0t = 0 to t=5t = 5 is split into 5 equal strips.

  3. Ordinate y0y_0 at t=0t = 0

    y0=40y_0 = 40

    Substitute t=0t = 0 into 40−t240 - t^2.

  4. Ordinate y1y_1 at t=1t = 1

    y1=39y_1 = 39

    Substitute t=1t = 1 into 40−t240 - t^2.

  5. Ordinate y2y_2 at t=2t = 2

    y2=36y_2 = 36

    Substitute t=2t = 2 into 40−t240 - t^2.

  6. Ordinate y3y_3 at t=3t = 3

    y3=31y_3 = 31

    Substitute t=3t = 3 into 40−t240 - t^2.

  7. Ordinate y4y_4 at t=4t = 4

    y4=24y_4 = 24

    Substitute t=4t = 4 into 40−t240 - t^2.

  8. Ordinate y5y_5 at t=5t = 5

    y5=15y_5 = 15

    Substitute t=5t = 5 into 40−t240 - t^2.

  9. Add the two end ordinates

    y0+y5=40+15=55y_0 + y_5 = 40 + 15 = 55

    The first and last ordinates are each used once.

  10. Add the middle ordinates

    39+36+31+24=13039 + 36 + 31 + 24 = 130

    Every ordinate strictly between the ends is used twice, so total them first.

  11. Substitute into the rule

    A≈12[40+15+2(39+36+31+24)]A \approx \frac{1}{2}\Big[40 + 15 + 2\big(39 + 36 + 31 + 24\big)\Big]

    Ends once each, middles twice each.

  12. Work out the estimate

    A≈12×315=157.5A \approx \frac{1}{2} \times 315 = 157.5

    The bracket totals 315; multiplying by half the strip width (0.5) gives the estimate.

  13. Is it an over- or under-estimate?

    under-estimate\text{under-estimate}

    The curve is concave here (it bends downwards), so every chord lies BELOW the curve. Each trapezium misses a sliver of area, and the estimate is an UNDER-estimate.

  14. How to improve the estimate

    more strips⇒smaller error\text{more strips} \Rightarrow \text{smaller error}

    Narrower strips hug the curve more closely, so the total sliver of missed or extra area shrinks and the estimate improves. The direction of the bias does not change.

  15. State the estimate

    A≈157.5 mA \approx 157.5\text{ m}

    The estimated distance is 157.5 m. The speed–time curve is concave, so the chords lie below it and the estimate is an under-estimate.

Answer
157.5 m157.5\text{ m}
Question 2
5 markschallenging
The exact area under the curve y=x3y = x^3 between x=0x = 0 and x=2x = 2 is 44. The trapezium rule with 44 strips of width 0.50.5 is used to estimate this area. Work out the percentage error in the estimate.
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Worked solution

  1. Find the strip width

    h=2−04=0.5h = \frac{2 - 0}{4} = 0.5

    Split the interval into 4 equal strips.

  2. List the ordinates

    y0=0,y1=0.125,y2=1,y3=3.375,y4=8y_0 = 0,\quad y_1 = 0.125,\quad y_2 = 1,\quad y_3 = 3.375,\quad y_4 = 8

    Substitute each x-value into x3x^3.

  3. Apply the trapezium rule

    A≈0.52[0+8+2(0.125+1+3.375)]A \approx \frac{0.5}{2}\Big[0 + 8 + 2\big(0.125 + 1 + 3.375\big)\Big]

    Ends once, middles twice, then multiply by half the strip width.

  4. Work out the estimate

    A≈0.52×17=4.25A \approx \frac{0.5}{2} \times 17 = 4.25

    This is the trapezium-rule estimate of the area.

  5. Write down the true area

    Atrue=4A_{\text{true}} = 4

    The exact area is given as 4, so it can be compared directly with the trapezium estimate.

  6. Find the error

    error=4.25−4=0.25\text{error} = 4.25 - 4 = 0.25

    The estimate is larger than the true area — an over-estimate, exactly as the convex shape predicts.

  7. Set up the percentage error

    % error=∣error∣true value×100\text{\% error} = \frac{|\text{error}|}{\text{true value}} \times 100

    Percentage error is always measured against the TRUE value, not the estimate.

  8. Substitute

    % error=0.254×100\text{\% error} = \frac{0.25}{4} \times 100

    Put the numbers in.

  9. Evaluate

    0.254×100=6.25%\frac{0.25}{4} \times 100 = 6.25\%

    Divide, then multiply by 100.

  10. Interpret the size of the error

    6.25% with 4 strips6.25\% \text{ with } 4 \text{ strips}

    A few per cent out is typical for so few strips; doubling the number of strips would cut the error sharply.

  11. Check the direction again

    over-estimate\text{over-estimate}

    Convex curve, chords above the curve, so the trapezia are too big. ✓

  12. Common error to avoid

    0.254.25×100=10017% (wrong denominator)\frac{0.25}{4.25} \times 100 = \frac{100}{17}\% \text{ (wrong denominator)}

    Dividing by the estimate instead of the true value gives a different, incorrect answer.

  13. Reflect

    estimate→error→% error\text{estimate} \to \text{error} \to \text{\% error}

    The three-step chain works for any estimation question, not just areas.

  14. Round sensibly

    ≈6.2%\approx 6.2\%

    One decimal place is plenty for a percentage error.

  15. State the percentage error

    6.25%6.25\%

    The 4-strip trapezium estimate of 4.25 is 6.25% larger than the true area of 4.

Answer
6.25%6.25\%
Question 3
5 markschallenging
A train's distance travelled, ss metres, after tt seconds is modelled by s=t2+4ts = t^2 + 4t. A tangent drawn to the curve at t=4t = 4 passes through (2, 8)(2,\ 8) and (6, 56)(6,\ 56). By how much does the train's speed at t=4t = 4 seconds exceed its average speed over the first 44 seconds?
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Worked solution

  1. Two different gradients are needed

    tangent gradient (instant)vschord gradient (average)\text{tangent gradient (instant)} \quad \text{vs} \quad \text{chord gradient (average)}

    The speed AT an instant is the gradient of the tangent; the AVERAGE speed over an interval is the gradient of the chord. They are different things.

  2. Locate the point on the curve at t=4t = 4

    y=32y = 32

    Substitute t=4t = 4 into t2+4tt^2 + 4t.

  3. Use the two points on the tangent

    (2, 8)and(6, 56)(2,\ 8) \quad \text{and} \quad (6,\ 56)

    These two points lie on the straight tangent line drawn at that point.

  4. Rise of the tangent

    rise=56−8=48\text{rise} = 56 - 8 = 48

    Subtract the vertical coordinates of the tangent points.

  5. Run of the tangent

    run=6−2=4\text{run} = 6 - 2 = 4

    Subtract the horizontal coordinates.

  6. Gradient of the tangent

    484=12\frac{48}{4} = 12

    So the instantaneous rate at t=4t = 4 is about 12 m/s12\text{ m/s}.

  7. Now the average: find the two end values

    t=0⇒y=0,t=4⇒y=32t = 0 \Rightarrow y = 0, \qquad t = 4 \Rightarrow y = 32

    Substitute both ends into t2+4tt^2 + 4t.

  8. Gradient of the chord

    32−04−0=324=8\frac{32 - 0}{4 - 0} = \frac{32}{4} = 8

    The average rate across the interval is 8 m/s8\text{ m/s}.

  9. Compare the two

    12and812 \quad \text{and} \quad 8

    The speed at t=4t = 4 s is 12 m/s, but the average speed over the first 4 seconds is only 8 m/s, because the train was slower earlier in the interval.

  10. Subtract

    12−8=412 - 8 = 4

    The question asks by how much the instantaneous value exceeds the average.

  11. Attach the units

    4 m/s4\text{ m/s}

    Both gradients are in the same units, so the difference is too.

  12. Why the two differ

    curve≠straight line\text{curve} \ne \text{straight line}

    On a straight-line graph the chord and the tangent are the same line, so the average and instantaneous rates agree. On a curve they do not.

  13. Both answers are estimates in an exam

    tangent drawn by hand\text{tangent drawn by hand}

    The chord gradient is exact (it uses two known points on the curve), but the tangent gradient depends on how well the tangent is drawn, so a small range is accepted.

  14. Check the tangent is steeper

    12>812 > 8

    A curve that bends upwards is steepest at the right-hand end of an interval, so the tangent there must be steeper than the chord across the whole interval. ✓

  15. State the answer

    4 m/s4\text{ m/s}

    The train's speed at t=4t = 4 exceeds its average speed over the first 4 seconds by 4 m/s.

Answer
4 m/s4\text{ m/s}
Question 4
5 markschallenging
A tangent is drawn to the curve y=x3−6x2+9xy = x^3 - 6x^2 + 9x at the point where x=1x = 1. The tangent passes through (0, 4)(0,\ 4) and (3, 4)(3,\ 4). Estimate the gradient of the curve at x=1x = 1.
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Worked solution

  1. What a tangent tells you

    gradient of curve at x=1≈gradient of tangent\text{gradient of curve at } x = 1 \approx \text{gradient of tangent}

    At GCSE you estimate the rate of change at a point by drawing a tangent there and finding the gradient of that straight line.

  2. Find the point on the curve

    x=1⇒y=4x = 1 \Rightarrow y = 4

    Substituting x=1x = 1 into x3−6x2+9xx^3 - 6x^2 + 9x gives the point where the tangent touches.

  3. Take the two points on the tangent

    (0, 4)and(3, 4)(0,\ 4) \quad \text{and} \quad (3,\ 4)

    Pick two points that lie on the straight tangent line and are far apart, so that small reading errors matter less.

  4. Find the rise

    rise=4−4=0\text{rise} = 4 - 4 = 0

    Subtract the vertical coordinates.

  5. Find the run

    run=3−0=3\text{run} = 3 - 0 = 3

    Subtract the horizontal coordinates.

  6. Divide rise by run

    gradient=03=0\text{gradient} = \frac{0}{3} = 0

    This is the gradient of the tangent, and so the estimated gradient of the curve.

  7. Interpret the gradient

    rate of change≈0\text{rate of change} \approx 0

    A gradient of zero means the tangent is horizontal — the curve is momentarily flat, so x=1x = 1 is a turning point of the curve.

  8. Why this is only an estimate

    hand-drawn tangent⇒estimate\text{hand-drawn tangent} \Rightarrow \text{estimate}

    Two students drawing the tangent by eye will get slightly different lines, so their answers will differ slightly. An exam accepts a small range around the true value.

  9. Check the sign against the shape

    0=00 = 0

    The curve is momentarily flat at this point, so the gradient must be zero. ✓

  10. Common error to avoid

    do not use two points on the CURVE\text{do not use two points on the CURVE}

    Reading two points off the curve gives a chord gradient (an average), not the gradient at the point. The two points must be on the tangent.

  11. Sanity check with a short chord

    y(0.9)=3.969,y(1)=4,y(1.1)=3.971y(0.9) = 3.969,\qquad y(1) = 4,\qquad y(1.1) = 3.971

    Both neighbouring values are just BELOW 4, so the curve rises to x=1x = 1 and then falls: x=1x = 1 is a local maximum. A short chord either side has a gradient near zero, confirming the tangent is flat.

  12. Record the units

    units:gradienthasnounitshere\text{units}: gradient has no units here

    Always attach the units the context demands.

  13. Reflect on the method

    tangent→gradient→rate\text{tangent} \to \text{gradient} \to \text{rate}

    Draw the tangent, pick two points far apart on it, divide rise by run — that is the whole GCSE method for the rate of change of a curve.

  14. Compare with an average rate

    tangent gradient≠chord gradient\text{tangent gradient} \ne \text{chord gradient}

    The tangent gradient measures the rate at one instant; a chord gradient would measure the average across an interval.

  15. State the estimate

    00

    The tangent is horizontal, so the estimated gradient at x=1x = 1 is 0 — the curve has a turning point there.

Answer
00
Question 5
6 markschallenging
A cyclist's speed, vv m/s, after tt seconds is modelled by v=t2v = t^2 for 0≤t≤40 \le t \le 4. Use the trapezium rule with 44 strips of width 11 to estimate the distance travelled, and hence estimate the cyclist's average speed over the 44 seconds.
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Worked solution

  1. Area under a speed–time graph is distance

    area under v–t graph=distance travelled\text{area under } v\text{–}t \text{ graph} = \text{distance travelled}

    Speed (m/s) multiplied by time (s) gives metres, so the area under the curve is a distance.

  2. Write down the trapezium rule

    A≈h2[y0+y4+2(y1+⋯+y3)]A \approx \frac{h}{2}\Big[y_0 + y_4 + 2\big(y_1 + \dots + y_{3}\big)\Big]

    Ends once each, middles twice each, all multiplied by half the strip width.

  3. Find the strip width

    h=4−04=1 sh = \frac{4 - 0}{4} = 1\text{ s}

    The 4 seconds are split into 4 equal strips.

  4. Speed at t=0t = 0 s

    y0=0 m/sy_0 = 0\text{ m/s}

    Substitute t=0t = 0 into t2t^2.

  5. Speed at t=1t = 1 s

    y1=1 m/sy_1 = 1\text{ m/s}

    Substitute t=1t = 1 into t2t^2.

  6. Speed at t=2t = 2 s

    y2=4 m/sy_2 = 4\text{ m/s}

    Substitute t=2t = 2 into t2t^2.

  7. Speed at t=3t = 3 s

    y3=9 m/sy_3 = 9\text{ m/s}

    Substitute t=3t = 3 into t2t^2.

  8. Speed at t=4t = 4 s

    y4=16 m/sy_4 = 16\text{ m/s}

    Substitute t=4t = 4 into t2t^2.

  9. Substitute into the rule

    A≈12[0+16+2(1+4+9)]A \approx \frac{1}{2}\Big[0 + 16 + 2\big(1 + 4 + 9\big)\Big]

    Ends once, middles twice.

  10. Work out the distance

    A≈12×44=22 mA \approx \frac{1}{2} \times 44 = 22\text{ m}

    The estimated distance travelled is about 22 metres.

  11. Is it an over- or under-estimate?

    over-estimate\text{over-estimate}

    The speed–time curve is convex (bending upwards), so each chord lies above the curve and the trapezia are slightly too big — an over-estimate.

  12. Now use average speed = distance ÷ time

    average speed=total distancetotal time\text{average speed} = \frac{\text{total distance}}{\text{total time}}

    This is the ordinary definition of average speed; the curve is no longer needed.

  13. Substitute

    average speed≈224\text{average speed} \approx \frac{22}{4}

    The estimated distance is 22 m and the time is 4 s.

  14. Evaluate

    224=5.5 m/s\frac{22}{4} = 5.5\text{ m/s}

    Divide to get the average speed.

  15. State the average speed

    5.5 m/s5.5\text{ m/s}

    The estimated distance is 22 m in 4 s, so the estimated average speed is 5.5 m/s.

Answer
5.5 m/s5.5\text{ m/s}

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